I belive we study them because in important categories they are close to free objects and even a retract of a free object in some algebraic instances (for example, direct summands in Mod_R, and precisely the free objects in Gps).
Am I right? Are there other reasons?

Seconding Theo B.'s comment: while one can define "universal delta-functor" without reference to any particular class of "resolutions" for computing, very quickly the mapping properties of projective/injectives (which yield chain homotopies proving independence of choice... etc.) become compellingly interesting for both computation and to prove existence of the universal delta-functors. These are the very simplest acyclic objects, being universally so (whenever proj/inj make sense). Even if we declare them somewhat auxiliary, they play a pivotal role.
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paul garrettJul 16 '11 at 19:44

I think your question is a little ambiguous: in an arbitrary category (and even abelian category) projective and injective objects play a perfectly dual role (i.e., upon switching to the opposite category). Thus the right answer to give in this level of generality is universal $\delta$-functors.
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Pete L. ClarkJul 16 '11 at 21:09

But on the other hand, in the category of modules over a ring, projective and injective objects behave very differently: e.g. free modules are necessarily projective but rarely injective. And indeed projective modules are interesting for lots of things other than the machinery of homological algebra. So which are you interested in, arbitrary categories or modules over a ring?
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Pete L. ClarkJul 16 '11 at 21:09

6 Answers
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I think the main reason for the interest in projective modules is the extraordinary bijective correspondence discovered by Serre between finitely generated projective modules $P$ on a completely arbitrary ring $A$ and locally free sheaves of finite rank (= vector bundles) $\mathcal F$ on the corresponding affine scheme $X=Spec(A)$. The correspondence simply associates to a vector bundle $\mathcal F$ the $A$-module of its global sections: $P=\Gamma(X, \mathcal F)$.

This gives a dictionary between algebra and geometry in which free modules correspond to trivial bundles, etc. Swan and Forster have perfected this dictionary by bringing topology and complex analysis onto the scene. A particularly interesting application of that dictionary is that one could prove that the projective module over the ring of functions of the 2-sphere corresponding to the tangent bundle of that sphere is not trivial by appealing to the well known topological fact that the tangent bundle is topologically non trivial. No proof had been available before.

Maybe the most spectacular succcess story of the dictionary is the simultaneous but independent proof by Quillen and Suslin in 1977 that every vector bundle on affine space $\mathbb A^n_k=Spec(k[T_1,\ldots,T_n])$ over a field $k$ is trivial by showing that finitely generated projective bundles over $k[T_1,\ldots,T_n]$ are trivial . This answered a question Serre had asked on page 243 of the article in which he introduced his correspondence: Faisceaux Algébriques Cohérents, always lovingly called FAC . This article is one of the most important ones in twentieth century mathematics. (English translation here, found thanks to MathOverflow)

Edit At Zev's request, let me try to explain why vector bundles give rise to projective modules . I'll do it in the more familiar context of topology, the idea being the same as in algebraic geometry.
Given a vector bundle $F$ on a compact topological space $X$, it is easy to see that there is a finite number of continuous sections $s_1, \ldots,s_N\in \Gamma(X,F)$ which generate the fiber of $F$ at each $x\in X$, even though $\Gamma(X,F)$ itself is (almost) always infinite dimensional.This yields a surjective morphism of vector bundles
$\pi : \theta^N\to F$ from the rank-$N$ trivial bundle on $X$ to $F$.
The kernel of $\pi$ is then a vector bundle $K$ on $X$ and the exact sequence of bundles $0\to K \to \theta^N \to F\to 0$ splits (they all do on compact spaces !) and we obtain a direct sum decomposition $\theta ^N=K\oplus F$. Taking global sections we obtain
$$C(X)^N= \Gamma(X,K) \oplus \Gamma(X,F)$$
from which projectivity ( and finite generation) of $ \Gamma(X,F) $ over $C(X)$ follows

+1 for your excellent answer. Are there illustrative examples of how non-projective modules go about failing to be locally free, or of finite rank, or both? Conversely, is there a geometric/intuitive explanation for why we should expect the ring of global sections of a vector bundle to always be projective? (I have not read Serre's or Swan's relevant papers, or indeed any proof of the correspondence; perhaps the answer is simply that it will be clear upon understanding the proof).
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Zev ChonolesJul 16 '11 at 19:23

Dear Zev, of course the easy way for me would be to agree that everything will become clear upon reading the proofs! But I'll try a less lazy solution and add a few words in an edit.
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Georges ElencwajgJul 16 '11 at 20:12

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But projective modules were studied and considered interesting quite a while before Serre proved his theorem?
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Mariano Suárez-Alvarez♦Jul 16 '11 at 20:36

@Mariano: this is true, but not by as much as I for one would have thought. Projective modules appeared on the scene surprisingly late: I believe they were introduced for the first time in the book of Cartan and Eilenberg. Serre's work on FAC was carried out at approximately the same time -- the 1950s -- and in close contact with the Athena-like emergence of homological algebra. But you're right that at least implicitly earlier roots can be found, e.g. ideal class groups and Picard groups.
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Pete L. ClarkJul 16 '11 at 21:15

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@Georges: According to Cartan (see second page) the book was completed by 1953: "Our initial project of a mere article for a journal was transformed; it became a book that we would propose to a publisher and for which it would be necessary to find a title that captured its content. We finally agreed on the term Homological Algebra. The text was given to Princeton University Press in 1953. I do not know why the book appeared only in 1956." @Pete: Georges replied to you in the comment above.
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t.b.Jul 17 '11 at 0:01

If I've understood Pete Clark's answer to another question correctly, here is a basic reason: in $R\text{-Mod}$ ($R$ a commutative ring) the dualizable objects are precisely the finitely-generated projective modules. For a good discussion of the importance of dualizable objects see, for example, Ponto and Shulman's Traces in symmetric monoidal categories.

This seems right to me to (although I haven't thought about it in that much generality). By the way, I think the idea of concentrating on finitely generated projective modules is a key one, as e.g. in the Serre-Swan theorem and the definition of $K_0$ of a ring. I would tentatively mark this as a separation between geometry and homological algebra.
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Pete L. ClarkJul 16 '11 at 21:18

Projective objects $P$ are important because they are the ones for which $\hom(P, -)$ is an exact functor - it is usually only half exact. Similarly, injective objects are the ones for which $\hom(-,I)$ is exact. This means that the derived functors (Ext-functors) of $\hom(P,-)$ and $\hom(-,I)$ vanish.

If you are given a ring, it is sometimes quite non-obvious how to construct its modules.

One natural place to look for modules is inside the ring itself and inside the free modules, which are immediately constructible from the ring itself. Now, inside free modules there can be all sort of things... but it is quite apparent that the simplest of those modules we are going to find inside free modules are precisely the summands thereof.

Devil's advocate: it's quite obvious how to construct the modules over a ring: they are all going to be quotients of free modules. Given that you get all modules this way, why would you want to look at only submodules of free modules?
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Pete L. ClarkJul 16 '11 at 23:04

@Pete: one reason: because it is very easy to manipulate elements of free modules—in particular, comparing them is quite simple. Comparisons in quotients, on the other hands, tend to be more elaborate.
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Mariano Suárez-Alvarez♦Jul 16 '11 at 23:24

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It is very illuminating to watch someone who does representation theory describe the categry of modules of, say, a finite dimensional hereditary algebra (think of a path algebra). Thy start from the modules you know are there, the projectives. Then they construct modules (using a process called knitting, due to Gabriel et al.) from there (they end up with the so called preprojective components of the AR quiver). Then do the same `backwards' from the injectives, which are just the projectives of something else. And then... you have to be very imaginative to come up with the other modules!
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Mariano Suárez-Alvarez♦Jul 16 '11 at 23:27

@Theo: the path algebroid (I would simply call it a $k$-linear category!) is Morita equivalent to the algebra, so in most senses related to representation theory, they are the same thing. One does deal with the algebroid in some contexts, though: for example, when one does covering theory (as in Gabriel et al.) one ends up considering quivers with infinitely many vertices, and then it is just simpler to adopt the language of $k$-linear categories. Nowadays, people do that, and look at modules over categories, and so on.
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Mariano Suárez-Alvarez♦Jul 17 '11 at 0:46

(1) In terms of representation theory, an $R$-module $M$ determines a representation of the ring $R$:

$R \rightarrow \text{End}(M)$

A projective $R$-module $P$ corresponds to a sub representation of a standard representation of $R$ on a free $R$-module $M = \prod_{i \in I} R$ that is also a direct summand: $M = P\oplus Q$. Thus, the map

Thus, once we know which free module $P$ is a direct summand of, we immediately know how $R$ acts on it.

(2) Linear functionals $P \rightarrow R$ on a projective $R$-module $P$ separate points, i.e. for all $p, q, \in P$, there exists a linear functional $f$ such that $f(p) \neq f(q)$. Because $f$ is linear, we can assume that $q = 0$, and so this is equivalent to the following: for every $p \in P$, there exists $f \in \text{Hom}_R(P, R)$, such that $f(p) \neq 0$. From this we see that this condition precisely says that the canonical map

$P \stackrel{\varphi}{\rightarrow} (P^\ast)^\ast$

$p \mapsto \varphi_p : (f \mapsto f(p))$

is injective, i.e. $\varphi_p = 0$, if, and only if, $f(p) = 0$ for all $f \in \text{Hom}_R(P, R)$, if, and only if, $p = 0$. Now, to show that $\varphi$ is injective, note that there is an $R$-module $Q$ such that $M = P\oplus Q$ is free. Suppose that $p\neq 0$. Let $f \in \text{Hom}_R(M, R)$, such that $f(p) \neq 0$, and restrict $f$ to $P$ (i.e. just take $f$ to be the projection onto a non-zero component of $p$). Then $\phi_p(f) \neq 0$. Thus, $\phi$ is injective.

More generally, an $R$-module $M$ is called torsionless if the canonical map $M \rightarrow (M^\ast)^\ast$, is injective. Thus projective modules are torsionless.